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Maksim231197 [3]
2 years ago
10

A company manufactures two types of trucks. Each truck must go through the painting shop and the assembly shop. If the painting

shop were completely devoted to painting type 1 trucks, 800 per day could be painted, whereas if the painting shop were completely devoted to painting type 2 trucks, 700 per day could be painted. If the assembly shop were completely devoted to assembling truck 1 engines, 1500 per day could be assembled, whereas if the assembly shop were completely devoted to assembling truck 2 engines, 1200 per day could be assembled. It is possible, however, to paint both types of trucks in the painting shop. Similarly, it is possible to assemble both types in the assembly shop. Each type 1 truck contributes $1000 to profit; each type 2 truck contributes $1500. Use Solver to maximize the company’s profit. (Hint: One approach, but not the only approach, is to try a graphical procedure first and then deduce the constraints from the graph.)
Mathematics
2 answers:
marin [14]2 years ago
8 0

Answer:

Step-by-step explanation:

Nastasia [14]2 years ago
6 0
Because bah. Njhg Biko Higgs yk
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A certain article indicates that in a sample of 1,000 dog owners, 610 said that they take more pictures of their dog than of the
lbvjy [14]

Answer:

(a) The 90% confidence interval is: (0.60, 0.63) Correct interpretation is (3).

(b) The 95% confidence interval is: (0.42, 0.46) Correct interpretation is (1).

(c) First, the confidence level in part(b) is <u>more than</u> the confidence level in part(a) is, so the critical value of part(b) is <u>more than</u> the critical value of part(a), Second the <u>margin of error</u> in part(b) is <u>more than</u> than in part(a).

Step-by-step explanation:

(a)

Let <em>X</em> = number of dog owners who take more pictures of their dog than of their significant others or friends.

Given:

<em>X</em> = 610

<em>n</em> = 1000

Confidence level = 90%

The (1 - <em>α</em>)% confidence interval for population proportion is:

CI=\hat p\pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

The sample proportion is:

\hat p=\frac{X}{n}=\frac{610}{1000}=0.61

The critical value of <em>z</em> for a 90% confidence level is:

z_{\alpha/2}=z_{0.10/2}=z_[0.05}=1.645

Construct a 90% confidence interval for the population proportion of dog owners who take more pictures of their dog than their significant others or friends as follows:

CI=\hat p\pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}\\=0.61\pm 1.645\sqrt{\frac{0.61(1-0.61}{1000}}\\=0.61\pm 0.015\\=(0.595, 0.625)\\\approx(0.60, 0.63)

Thus, the 90% confidence interval for the population proportion of dog owners who take more pictures of their dog than their significant others or friends is (0.60, 0.63).

<u>Interpretation</u>:

There is a 90% chance that the true proportion of dog owners who take more pictures of their dog than of their significant others or friends falls within the interval (0.60, 0.63).

Correct option is (3).

(b)

Let <em>X</em> = number of dog owners who are more likely to complain to their dog than to a friend.

Given:

<em>X</em> = 440

<em>n</em> = 1000

Confidence level = 95%

The (1 - <em>α</em>)% confidence interval for population proportion is:

CI=\hat p\pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

The sample proportion is:

\hat p=\frac{X}{n}=\frac{440}{1000}=0.44

The critical value of <em>z</em> for a 95% confidence level is:

z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96

Construct a 95% confidence interval for the population proportion of dog owners who are more likely to complain to their dog than to a friend as follows:

CI=\hat p\pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}\\=0.44\pm 1.96\sqrt{\frac{0.44(1-0.44}{1000}}\\=0.44\pm 0.016\\=(0.424, 0.456)\\\approx(0.42, 0.46)

Thus, the 95% confidence interval for the population proportion of dog owners who are more likely to complain to their dog than to a friend  is (0.42, 0.46).

<u>Interpretation</u>:

There is a 95% chance that the true proportion of dog owners who are more likely to complain to their dog than to a friend falls within the interval (0.42, 0.46).

Correct option is (1).

(c)

The confidence interval in part (b) is wider than the confidence interval in part (a).

The width of the interval is affected by:

  1. The confidence level
  2. Sample size
  3. Standard deviation.

The confidence level in part (b) is more than that in part (a).

Because of this the critical value of <em>z</em> in part (b) is more than that in part (a).

Also the margin of error in part (b) is 0.016 which is more than the margin of error in part (a), 0.015.

First, the confidence level in part(b) is <u>more than</u> the confidence level in part(a) is, so the critical value of part(b) is <u>more than</u> the critical value of part(a), Second the <u>margin of error</u> in part(b) is <u>more than</u> than in part(a).

4 0
2 years ago
The number of newly reported crime cases in a county in New York State is shown in the accompanying table, where x represents th
Novay_Z [31]

Answer:

Step-by-step explanation:

I use  84+ CE

stat edit, then fill in the #s

then

vars 5

then

2'nd stat plot, on

then, click stat

Click arrow 1 time to the left to get to Calc

then click (4)(LinReg(ax+b))

then click enter 5 times

(y=-25.31428571x+1000.285714

y=-25.3x+1000.3

now, lets use computer:

y=-25.31(543)+1000.3

y=-12743.03

round to the biggest whole number )

this doesn't really work, so I will put 1999, 2000, 2001, 2002, 2003, 2004 instead of 0, 1, 2, 3, 4, 5 and do the same thing

now I get

y=-25.31428571x+51603.54286

y=-25.3x+51603.5

now, lets use computer:

y=-25.3(543)+51603.5

y=37865.6

round to the biggest whole number:

y=37866

so, year 37866

7 0
2 years ago
The weight, y, in pounds, of kittens was tracked for the first 8 weeks after birth where t represents the number of weeks after
marshall27 [118]

Answer:

Step-by-step explanation:

The mentioned relationship for the weight, in pounds, of the kitten with respect to time, in weeks, is

\hat y =1.7 +1.48t

Weight of the kitten after 10 weeks

\hat y =1.7 +1.48\times 10

\hat y =16.5 pounds

This modeled equation is based on the observation of the early age of a kitten where the kitten is in its growth period, but in the early stage the growth rate in the weight of the kitten was the same but the growth of any living beings continues till the adult stage. So, after some time, in real life situation, this weekly change in weight will become zero, So, this model is not suitable to measure the weight of the kitten over the larger time period.

Here, t= 10 weeks is nearby the observed time period, so the linearly modeled equation can be used to predict the weight.

Hence, the weight of the kitten after 10 weeks is 16.5 pounds.

8 0
2 years ago
The Hazell family has 4 children. Murphy is 1 year younger than his older brother Michael. Keira is 2 years younger than Murphy.
Allushta [10]
Isabelle is 7 years old
6 0
2 years ago
Computer upgrades have a nominal time of 80 minutes. samples of five observations each have been taken, and the results are as l
Goshia [24]
The answer:
<span>the upper and lower control limits (uclim and lclim) for mean formula is 

for the mean chart
uclim= x+A2xR
where x = sum(of the value)  / number of each value
and for 
lclim=</span>x+A2xR
<span>
R is the range such that R= Xmax - Xmin

in the case of the sample 1: S1
the data are: 
79.2    78.8   80.0   78.4   81.0

the mean is   x1 = (</span>79.2  +  78.8  +  80.0  + 78.4  + 81.0)  / 5=  79.48
<span>its range is    R 1= 81.0 -78.4 = 2.6

we can do the same method for finding the mean chart and range for all samples
</span>S2: x2=<span> 80.14  ,  R2=2.3
</span>S3: x3= 80.14  ,  R3=1.2
S4: x4= 79.60  ,  R4=1.7
S5: x5= 80.02  ,  R5=2.0
S6: x6=80.38   ,  R6=1.4
<span>
therefore the average value is   X= sum( x1+x2+...+x6) / 6 = 79.96
and R=sum(R1+R2+...+R6)/6=1.87
finally
range chart uclim =D4xR=3.95  and lclim is always equal to 0, because D3=0

we can say that the process is not in control.
       




</span>
3 0
2 years ago
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